# Why isn't $\sqrt{64x^4y^8z^6}$ equal to $8x^2y^4z^3$?

I simplified $$\sqrt{64x^4y^8z^6}$$ by taking the square root of $$64$$ (getting $$8$$), $$x^4$$ (getting $$x^2$$), $$y^8$$ (getting $$y^4$$), and $$z^6$$ (getting $$z^3$$). My answer, $$8x^2y^4z^3$$, not quite right and I am having a hard time understanding why not.

I even placed the problem into WolframAlpha.com (here) and got the same answer, which was really confusing. Any pointers of what I might be doing wrong?

As per the hint, you have to be careful since $$z^3$$ is negative if $$z<0$$ and the square root refers to the principal (nonnegative) root. Wolfram also mentions the assumption that $$x,y,z$$ are positive in their simplification to your answer. But for arbitrary real $$x,y,z$$ we have

$$\sqrt{64x^4y^8z^6}=8x^2y^4|z|^3.$$

• Ahhhh, yes. I didn't realize that one part of the answer could have an absolute value specifically for it. Thank you for that. Commented Mar 9, 2022 at 6:23
• @יהודה no prob, feel free to accept the answer if it helps Commented Mar 9, 2022 at 6:24
• I was trying to accept but it said I had to wait 5 mins lol Commented Mar 9, 2022 at 6:26

It’s conventional for $$\sqrt n$$ to give you the positive square root of $$n$$, but in this case they might be expecting you to give the answer in the form $$\pm {8x^2y^4z^3}$$

• Thank you for that. Sheesh, so many different ways to answer a question. Fascinating actually. Commented Mar 9, 2022 at 6:28
• You probably want that without the \sqrt part - it happened already. Commented Mar 9, 2022 at 16:36